\(\int \frac {(c+d x) (a x^2+b x^3)^{3/2}}{x^6} \, dx\) [254]

Optimal result

Integrand size = 24, antiderivative size = 120 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=-\frac {(3 b c+4 a d) \sqrt {a x^2+b x^3}}{4 x^2}+\frac {2 b d \sqrt {a x^2+b x^3}}{x}-\frac {c \left (a x^2+b x^3\right )^{3/2}}{2 x^5}-\frac {3 b (b c+4 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{4 \sqrt {a}} \] Output:

-1/4*(4*a*d+3*b*c)*(b*x^3+a*x^2)^(1/2)/x^2+2*b*d*(b*x^3+a*x^2)^(1/2)/x-1/2 
*c*(b*x^3+a*x^2)^(3/2)/x^5-3/4*b*(4*a*d+b*c)*arctanh((b*x^3+a*x^2)^(1/2)/a 
^(1/2)/x)/a^(1/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.84 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=-\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} (b x (5 c-8 d x)+2 a (c+2 d x))+3 b (b c+4 a d) x^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 \sqrt {a} x^3 \sqrt {a+b x}} \] Input:

Integrate[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^6,x]
 

Output:

-1/4*(Sqrt[x^2*(a + b*x)]*(Sqrt[a]*Sqrt[a + b*x]*(b*x*(5*c - 8*d*x) + 2*a* 
(c + 2*d*x)) + 3*b*(b*c + 4*a*d)*x^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(Sqr 
t[a]*x^3*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1944, 1926, 1927, 1914, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (a x^2+b x^3\right )^{3/2} (c+d x)}{x^6} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(4 a d+b c) \int \frac {\left (b x^3+a x^2\right )^{3/2}}{x^5}dx}{4 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{2 a x^7}\)

\(\Big \downarrow \) 1926

\(\displaystyle \frac {(4 a d+b c) \left (\frac {3}{2} b \int \frac {\sqrt {b x^3+a x^2}}{x^2}dx-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )}{4 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{2 a x^7}\)

\(\Big \downarrow \) 1927

\(\displaystyle \frac {(4 a d+b c) \left (\frac {3}{2} b \left (a \int \frac {1}{\sqrt {b x^3+a x^2}}dx+\frac {2 \sqrt {a x^2+b x^3}}{x}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )}{4 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{2 a x^7}\)

\(\Big \downarrow \) 1914

\(\displaystyle \frac {(4 a d+b c) \left (\frac {3}{2} b \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 a \int \frac {1}{1-\frac {a x^2}{b x^3+a x^2}}d\frac {x}{\sqrt {b x^3+a x^2}}\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right )}{4 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{2 a x^7}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (\frac {3}{2} b \left (\frac {2 \sqrt {a x^2+b x^3}}{x}-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )\right )-\frac {\left (a x^2+b x^3\right )^{3/2}}{x^4}\right ) (4 a d+b c)}{4 a}-\frac {c \left (a x^2+b x^3\right )^{5/2}}{2 a x^7}\)

Input:

Int[((c + d*x)*(a*x^2 + b*x^3)^(3/2))/x^6,x]
 

Output:

-1/2*(c*(a*x^2 + b*x^3)^(5/2))/(a*x^7) + ((b*c + 4*a*d)*(-((a*x^2 + b*x^3) 
^(3/2)/x^4) + (3*b*((2*Sqrt[a*x^2 + b*x^3])/x - 2*Sqrt[a]*ArcTanh[(Sqrt[a] 
*x)/Sqrt[a*x^2 + b*x^3]]))/2))/(4*a)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1914

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[2/(2 - n) 
Subst[Int[1/(1 - a*x^2), x], x, x/Sqrt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, 
n}, x] && NeQ[n, 2]
 

rule 1926

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] 
 :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + j*p + 1))), x] - Simp[b*p 
*((n - j)/(c^n*(m + j*p + 1)))   Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p - 1), 
 x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Integer 
sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
 

rule 1927

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[(c*x)^(m + 1)*((a*x^j + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a* 
(n - j)*(p/(c^j*(m + n*p + 1)))   Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p - 1) 
, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (Int 
egersQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && NeQ[m + n*p + 1, 0]
 

rule 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {\left (4 a d x +5 c b x +2 a c \right ) \sqrt {x^{2} \left (b x +a \right )}}{4 x^{3}}+\frac {b \left (16 \sqrt {b x +a}\, d -\frac {2 \left (12 a d +3 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{8 x \sqrt {b x +a}}\)	\(94\)
pseudoelliptic	\(-\frac {3 \left (b^{4} x^{5} \left (a d -\frac {b c}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {4 \left (-\frac {5 b^{3} x^{3} \left (3 d x +c \right ) a^{\frac {3}{2}}}{4}+b^{2} x^{2} \left (\frac {5 d x}{2}+c \right ) a^{\frac {5}{2}}+22 x b \left (\frac {15 d x}{11}+c \right ) a^{\frac {7}{2}}+\left (20 d x +16 c \right ) a^{\frac {9}{2}}+\frac {15 \sqrt {a}\, b^{4} c \,x^{4}}{8}\right ) \sqrt {b x +a}}{15}\right )}{64 a^{\frac {7}{2}} x^{5}}\)	\(117\)
default	\(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (-8 \sqrt {b x +a}\, d \,b^{2} x^{2} \sqrt {a}+12 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{2} d \,x^{2}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{3} c \,x^{2}+4 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {3}{2}} d +5 \left (b x +a \right )^{\frac {3}{2}} \sqrt {a}\, b c -4 \sqrt {b x +a}\, a^{\frac {5}{2}} d -3 \sqrt {b x +a}\, a^{\frac {3}{2}} b c \right )}{4 b \,x^{5} \left (b x +a \right )^{\frac {3}{2}} \sqrt {a}}\)	\(149\)

Input:

int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^6,x,method=_RETURNVERBOSE)
 

Output:

-1/4*(4*a*d*x+5*b*c*x+2*a*c)/x^3*(x^2*(b*x+a))^(1/2)+1/8*b*(16*(b*x+a)^(1/ 
2)*d-2*(12*a*d+3*b*c)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))*(x^2*(b*x+a) 
)^(1/2)/x/(b*x+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.77 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\left [\frac {3 \, {\left (b^{2} c + 4 \, a b d\right )} \sqrt {a} x^{3} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (8 \, a b d x^{2} - 2 \, a^{2} c - {\left (5 \, a b c + 4 \, a^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{8 \, a x^{3}}, \frac {3 \, {\left (b^{2} c + 4 \, a b d\right )} \sqrt {-a} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{b x^{2} + a x}\right ) + {\left (8 \, a b d x^{2} - 2 \, a^{2} c - {\left (5 \, a b c + 4 \, a^{2} d\right )} x\right )} \sqrt {b x^{3} + a x^{2}}}{4 \, a x^{3}}\right ] \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="fricas")
 

Output:

[1/8*(3*(b^2*c + 4*a*b*d)*sqrt(a)*x^3*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + 
a*x^2)*sqrt(a))/x^2) + 2*(8*a*b*d*x^2 - 2*a^2*c - (5*a*b*c + 4*a^2*d)*x)*s 
qrt(b*x^3 + a*x^2))/(a*x^3), 1/4*(3*(b^2*c + 4*a*b*d)*sqrt(-a)*x^3*arctan( 
sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) + (8*a*b*d*x^2 - 2*a^2*c - (5* 
a*b*c + 4*a^2*d)*x)*sqrt(b*x^3 + a*x^2))/(a*x^3)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}} \left (c + d x\right )}{x^{6}}\, dx \] Input:

integrate((d*x+c)*(b*x**3+a*x**2)**(3/2)/x**6,x)
 

Output:

Integral((x**2*(a + b*x))**(3/2)*(c + d*x)/x**6, x)
 

Maxima [F]

\[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}} {\left (d x + c\right )}}{x^{6}} \,d x } \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="maxima")
 

Output:

integrate((b*x^3 + a*x^2)^(3/2)*(d*x + c)/x^6, x)
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.11 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\frac {8 \, \sqrt {b x + a} b^{2} d \mathrm {sgn}\left (x\right ) + \frac {3 \, {\left (b^{3} c \mathrm {sgn}\left (x\right ) + 4 \, a b^{2} d \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} c \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {b x + a} a b^{3} c \mathrm {sgn}\left (x\right ) + 4 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{2} d \mathrm {sgn}\left (x\right ) - 4 \, \sqrt {b x + a} a^{2} b^{2} d \mathrm {sgn}\left (x\right )}{b^{2} x^{2}}}{4 \, b} \] Input:

integrate((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="giac")
 

Output:

1/4*(8*sqrt(b*x + a)*b^2*d*sgn(x) + 3*(b^3*c*sgn(x) + 4*a*b^2*d*sgn(x))*ar 
ctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - (5*(b*x + a)^(3/2)*b^3*c*sgn(x) - 
3*sqrt(b*x + a)*a*b^3*c*sgn(x) + 4*(b*x + a)^(3/2)*a*b^2*d*sgn(x) - 4*sqrt 
(b*x + a)*a^2*b^2*d*sgn(x))/(b^2*x^2))/b
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}\,\left (c+d\,x\right )}{x^6} \,d x \] Input:

int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/x^6,x)
 

Output:

int(((a*x^2 + b*x^3)^(3/2)*(c + d*x))/x^6, x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.22 \[ \int \frac {(c+d x) \left (a x^2+b x^3\right )^{3/2}}{x^6} \, dx=\frac {-4 \sqrt {b x +a}\, a^{2} c -8 \sqrt {b x +a}\, a^{2} d x -10 \sqrt {b x +a}\, a b c x +16 \sqrt {b x +a}\, a b d \,x^{2}+12 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b d \,x^{2}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{2} c \,x^{2}-12 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b d \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{2} c \,x^{2}}{8 a \,x^{2}} \] Input:

int((d*x+c)*(b*x^3+a*x^2)^(3/2)/x^6,x)
 

Output:

( - 4*sqrt(a + b*x)*a**2*c - 8*sqrt(a + b*x)*a**2*d*x - 10*sqrt(a + b*x)*a 
*b*c*x + 16*sqrt(a + b*x)*a*b*d*x**2 + 12*sqrt(a)*log(sqrt(a + b*x) - sqrt 
(a))*a*b*d*x**2 + 3*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**2*c*x**2 - 12* 
sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*a*b*d*x**2 - 3*sqrt(a)*log(sqrt(a + b 
*x) + sqrt(a))*b**2*c*x**2)/(8*a*x**2)